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Mathematics 332: Algebra – Midterm MATHEMATICS 332: ALGEBRA { MIDTERM 1. Introduction This exam is meant to help you work with various ideas from group theory in one context. As we have discussed, this exam is open instructor, and you are expected to check with me regularly on your progress, first in solving the problems and then second in expounding your solutions. Definition 1.1 (Modular Group). The modular group is the subgroup of SL2(R) consisting of the 2-by-2 matrices having integer entries and determinant 1, a b SL ( ) = : a; b; c; d 2 ; ad − bc = 1 : 2 Z c d Z Exercise 1. Verify that indeed SL2(Z) is a subgroup of SL2(R). 2. Principal Congruence Subgroups Let N be a positive integer. The reduce mod N map Z −! Z=NZ is a ring homomorphism that takes 1Z to 1Z=NZ. Consequently, applying the map entrywise to 2-by-2 matrices gives a group homomorphism SL2(Z) −! SL2(Z=NZ): Definition 2.1 (Principal Congruence Subgroup). Let N be a positive integer. The principal congruence subgroup of level N is the kernel of the entrywise reduction mod N map on SL2(Z), Γ(N) = ker SL2(Z) −! SL2(Z=NZ) : Thus Γ(N) is normal in SL2(Z). Specifically, a b a b 1 0 Γ(N) = 2 SL ( ): = mod N : c d 2 Z c d 0 1 (The matrix congruence is interpreted entrywise, i.e., a = 1 mod N and so on.) In particular Γ(1) = SL2(Z). The next exercise shows that the entrywise reduc- tion map SL2(Z) −! SL2(Z=NZ) is a surjection. a b Exercise 2. Let γ 2 SL2(Z=NZ) be given. Lift γ to a matrix c d 2 M2(Z). This exercise explains how to modify the lift so that its determinant is 1. (a) Show that gcd(c; d; N) = 1. (b) Assume that c 6= 0. Show that gcd(c; d0) = 1 for some d0 = d + tN where t 2 Z. (If c = 0 then d 6= 0|unless N = 1, in which case the whole problem is 1 2 MATHEMATICS 332: ALGEBRA { MIDTERM trivial|and a similar argument works, so we omit it.) Hint: Let t = Q p, pjc; p-d and show that if p j c then d + tN 6= 0 mod p. a+kN b+`N (c) Still working with the case c 6= 0, show that some lift c d0 of γ lies in SL2(Z). Thus the map surjects. (Again the c = 0 case is handled similarly, so we omit it. But note that the process of adjusting the lift to make its determinant equal 1 has involved all four of its entries. Since SL2(Z) is infinite and SL2(Z=NZ) is finite, perhaps the amount of work necessary to show that SL2(Z) −! SL2(Z=NZ) surjects is surprising.) Now that the homomorphism SL2(Z) −! SL2(Z=NZ) is known to surject, we have an isomorphism ∼ SL2(Z)=Γ(N) −! SL2(Z=NZ): Consequently the index [SL2(Z) : Γ(N)] is finite for all N. The next exercise is to show that specifically the index is Y 1 [SL ( ) : Γ(N)] = N 3 1 − ; 2 Z p2 pjN where the product is taken over all prime divisors of N. Exercise 3. (a) Let p be a prime and let e be a positive integer. Show by induction e 3e 2 on e that jSL2(Z=p Z)j = p (1 − 1=p ). (b) Cite the Sun-Ze Theorem and use one more idea to show that jSL2(Z=NZ)j = 3 Q 2 N pjN (1 − 1=p ), so this is also the index [SL2(Z) : Γ(N)]. 3. Congruence Subgroups in General Definition 3.1 (Congruence Subgroup, Level). Let N be a positive integer. A subgroup of SL2(Z) is a congruence subgroup of level N if it contains the principal congruence subgroup Γ(N). Equivalently, a subgroup of SL2(Z) is a con- gruence subgroup if it is the inverse image under reduction modulo N of a subgroup of SL2(Z=NZ). Since every Γ(N) has finite index in SL2(Z), so does every congruence sub- group Γ. Besides the principal congruence subgroups, the most important congru- ence subgroups are the inverse images of the SL2(Z=NZ)-subgroups 1 ∗ G = 2 SL ( =N ) 1 0 1 2 Z Z (where \∗" means \unspecified”) and ∗ ∗ G = 2 SL ( =N ) : 0 0 ∗ 2 Z Z Specifically, the congruence subgroups are a b a b 1 ∗ Γ (N) = 2 SL ( ): = mod N 1 c d 2 Z c d 0 1 and a b a b ∗ ∗ Γ (N) = 2 SL ( ): = mod N : 0 c d 2 Z c d 0 ∗ MATHEMATICS 332: ALGEBRA { MIDTERM 3 Thus for any positive integer N we have the chain of containments Γ(N) ⊂ Γ1(N) ⊂ Γ0(N) ⊂ SL2(Z): Exercise 4. (a) Show that the map a b g :Γ (N) −! =N ; 7−! b mod N 1 1 Z Z c d is an epimorphism with kernel Γ(N). (b) Show that the map a b g :Γ (N) −! ( =N )×; 7−! d mod N 0 0 Z Z c d is an epimorphism with kernel Γ1(N). It follows from exercise 4(a) that ∼ Γ(N) C Γ1(N); Γ1(N)=Γ(N) −! Z=NZ; [Γ1(N) : Γ(N)] = N: And it follows from exercise 4(b) that ∼ × Γ1(N) C Γ0(N); Γ0(N)=Γ1(N) −! (Z=NZ) ; [Γ0(N):Γ1(N)] = '(N); where ' is the Euler totient function. Exercise 4. (c) Show that in consequence of the indices in the two previous displays and of the previously-computed value of [SL2(Z) : Γ(N)], it follows that Y [SL2(Z):Γ0(N)] = N (1 + 1=p); pjN the product taken over all primes dividing N. 4. The Theta Group as a Congruence Subgroup Here is a sketched argument that the modular group is generated by the two matrices 1 1 0 −1 and : 0 1 1 0 Let Γ be the subgroup of SL2(Z) generated by the two matrices. Note that 1 n 1 1 n a b [ 0 1 ] = [ 0 1 ] 2 Γ for all n 2 Z. Let α = c d be a matrix in SL2(Z). The identity a b 1 n a b0 = c d 0 1 c nc + d shows that unless c = 0, some matrix αγ with γ 2 Γ has bottom row (c; d0) with jd0j ≤ jcj=2. The identity a b 0 −1 b −a = c d 1 0 d −c shows that this process can be iterated until some matrix αγ with γ 2 Γ has bottom row (0; ∗). Because we are working with matrices that have determinant 1, in fact 0 −1 2 the bottom row of αγ is (0; ±1), and since 1 0 = −I it can be taken to be 1 n (0; 1). It follows that αγ = [ 0 1 ] for some n 2 Z and hence that αγ 2 Γ. Thus α 2 Γ, and Γ is all of SL2(Z). 4 MATHEMATICS 332: ALGEBRA { MIDTERM Make sure that you understand this argument. If it gives you trouble then work with me in person or via email until you get it. Definition 4.1 (Theta Group). The theta group Γθ is the subgroup of SL2(Z) generated by the matrices 1 1 1 0 ± and ± : 0 1 4 1 Exercise 5. This exercise shows that Γθ = Γ0(4). The containment \⊂" holds because the four generators of Γθ lie in Γ0(4). For the other containment, let a b α = c d be a matrix in Γ0(4). Similarly to the discussion above, the identity a b 1 n a b0 = c d 0 1 c nc + d 0 shows that unless c = 0, some matrix αγ with γ 2 Γθ has bottom row (c; d ) with jd0j < jcj=2, but now the inequality is strict. Explain why the inequality is strict. Use the identity a b 1 0 a0 b = c d 4n 1 c + 4nd d 0 0 to show that unless d = 0, some matrix αγ with γ 2 Γθ has bottom row (c ; d ) with jc0j < 2jd0j. Again explain why the inequality is strict. Each multiplication reduces the positive integer quantity minfjcj; 2jdjg, so the process must stop with c = 0 or d = 0. Show that in fact this means that αγ 2 Γθ for some γ 2 Γθ and so α 2 Γθ..
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